# Show that a sequence is monotonically increasing.

It's given that $$a \in \mathbb{R}$$ with $$a > 0$$ and $$0. The sequence $$(x_n)_{n\in\mathbb{N}}$$ is defined by $$x_{n+1} = 2x_n-ax^2_n.$$ Now, I first proved by induction that $$x_n < \frac{1}{n}$$ for all $$n \in \mathbb{N}$$. Then I was wondering, if it is enough to just show that $$x_{n+1} = 2x_n-ax^2_n > 2x_n - \frac{1}{x_n}x^2_n = 2x_n-x_n = x_n$$ to fully prove that $$x_{n+1} > x_n$$ for all $$n \in \mathbb{N}$$ and therefore $$(x_n)_{n\in\mathbb{N}}$$ is monotonically increasing? Thanks for any corrections or tips, I'm just not sure if I might be missing something.

• Yes, it would be enough, but how do you prove it? – Bernard Jul 10 '19 at 22:58
• How do you know that $\frac1{x_n}\gt a$? – Chris Custer Jul 10 '19 at 22:59
• 1.: $x_1 < \frac{1}{a}$ by Definition. 2.: Let's suppose that $x_n < \frac{1}{a}$ is true. Then $x_{n+1} < \frac{1}{a}$ should follow. 3.: $x_{n+1} = 2x_n-ax^2_n = x_n(2-ax_n) < \frac{1}{a}(2-\frac{1}{a}a) = \frac{1}{a}.$ – psyph Jul 10 '19 at 23:05
• I don't think your induction step works out, because the second $x_n$ is replaced by the bigger $\frac1a$, and then subtracted. – Chris Custer Jul 10 '19 at 23:13
• Your proof is OK but you have to justify the inequality $x_n(2-ax_n) <\frac 1 a (2-\frac 1 a a)$. – Kavi Rama Murthy Jul 10 '19 at 23:14

You have to justify the inequality $$x_n(2-ax_n) <\frac 1 a (2-\frac 1 a a)$$. To prove this consider the function $$f(t)=t(2-at)$$ defined on $$(0,\frac 1 a)$$. Since $$f'(t)=2-2at>0$$ for all $$t$$ in the domain of $$f$$ this function is increasing. Since $$t(2-at)=\frac 1 a$$ when $$t=\frac 1 a$$ it follows (by looking at the extension of $$f$$ to $$(0,\frac 1 a]$$) that $$f(t) <\frac 1 a$$ for all $$t$$. So if we assume that $$x_n <\frac 1 a$$ we can take $$t=x_n$$ to see that $$x_{n+1}<\frac 1 a$$.
• Thanks, that made it so clear. Now can I also ask you, if I want to show that my sequence converges to $\frac{1}{a}$, can I make my proof by showing that $lim_{n\to\infty} (x_{n+1}-\frac{1}{a})=0$? – psyph Jul 10 '19 at 23:41
• The sequence is increasing and bounded above by $\frac 1 a$. This implies that it is convergent. If you call the limit $l$ then the given equation gives $l=2l-al^{2}$ or $l=al^{2}$. This means $l=0$ or $l =\frac 1 a$. Since $x_n >0$ (proved again by induction) for all $n$ and $x_n$ is increasing its limit cannot be $0$. Hence $l =\frac 1 a$. – Kavi Rama Murthy Jul 10 '19 at 23:49
• Thanks, but which equations gives $l=2l-al^2$ or $al^2$? And what about the proof that I proposed? Is it sufficient? Because it works out since $lim_{n\to\infty}(x_{n+1}-\frac{1}{a})$ is equal to zero. – psyph Jul 10 '19 at 23:54
• How do you show that $x_{n+1}-\frac 1 a \to 0$? – Kavi Rama Murthy Jul 10 '19 at 23:58
• I am just taking limit as $n \to \infty$ in the equation $x_{n+1}=2x_n-ax_n^{2}$. This gives $l=2l-al^{2}$. We can re-write this equation as $al^{2}=l$. This gives $l(al-1)=0$. So either $l=0$ or $al=1$. Once you rule out $l=0$ we are left with $al=1$ which gives $l=\frac 1 a$. – Kavi Rama Murthy Jul 11 '19 at 0:09